Prove that $\varphi$ is continuous on the real numbers and $\varphi(x) = x$ 
Let $\varphi : \mathbb{R} \to \mathbb{R}$ be a bijection such that $\varphi(x) = x\;\forall x \in\mathbb{Q}$ and $\varphi(xy) = \varphi(x)\varphi(y)\wedge \varphi(x+y) = \varphi(x)+\varphi(y)\forall x,y\in\mathbb{R}$. Prove that $\varphi(x)=x\;\forall x\in\mathbb{R}$.

Just to make sure I'm understanding everything right, I'm going to try to explain as much as I can in the following solution: Let $\varphi :\mathbb{R}\to\mathbb{R}$ be the said bijection. Then $\forall x\in\mathbb{R}$ such that $x>0$, $\exists y\in \mathbb{R}$ such that $x=y^2\Rightarrow \varphi(x)=\varphi(y^2)=[\varphi(y)]^2>0$. Since $x$ was arbitrary, $\varphi(x)$ preserves $>$. Between any two distinct real numbers there is a rational number so for all distinct $a,b\in\mathbb{R},\exists r\in\mathbb{Q}$ such that $a<r<b$. We thus have that $\varphi(a) < \varphi(r)=r<\varphi(b)$. By the Archimedean property of the real numbers, $\forall\epsilon >0$, there exists $a,b\in\mathbb{R}$ and $n\in\mathbb{N}$ such that $|a-b|\leq\frac{1}{10^n}<\epsilon$. I think this is done, but I want to use the fact that between any two real numbers there is a rational number. 
 A: Your proof as currently stated is not valid because your claim, as currently stated, is false. First, let me amend your claim. Did you actually mean this?

Let $\varphi : \mathbb{R} \to \mathbb{R}$ be a bijection such that $\forall x \in \mathbb{Q},  \varphi(x) = x$, $\forall x, y \in \mathbb{R}, \varphi(x+y) = \varphi(x) + \varphi(y)$, and $\forall x,y \in \mathbb{R}, \varphi(xy) = \varphi(x)\varphi(y)$. Then $\forall x \in \mathbb{R}, \varphi(x)=x$.

In that case, where I've added an extra condition about addition, your proof along with @Milten's works. The missing step in your proof is as follows.
You prove that if $x > 0$, then $\varphi(x) > 0$. This does not mean that $\varphi$ preserves order: it only means that it takes positive numbers to positive numbers and negative numbers to negative numbers. However, if you add my linearity (or technically, additivity) condition to your problem, then $\varphi$ does preserve order as follows:
Suppose $x > y$. Then $x - y > 0$, and so $\varphi(x-y) > 0$, and thus by linearity, $\varphi(x) - \varphi(y) > 0$, and so $\varphi(x) > \varphi(y)$, i.e., $\varphi$ preserves order. Do you see why my condition was necessary to make this work? Otherwise, there are lots of bijections of $\mathbb{R}$ out there that preserve being positive but don't preserve order. 
Anyway, if you add this condition, then the claim is true, and the proofs above suffice. 
What if you don't add the condition? Well, it turns out that your claim is then false, although this is much more difficult to prove, and since you're just starting out in Calculus and Linear Algebra, you probably don't have the vocabulary yet to understand why this is true, but I'll lay some of it out for you (and other users) here. 
I'm going to build a counterexample, i.e., a function $\varphi: \mathbb{R} \to \mathbb{R}$ that is a bijection, is constant on all rational numbers, and distributes over multiplication, but is not the identity function. This will disprove your claim. The construction uses something called the Axiom of Choice, which should suggest to you two things: this function will be described very indirectly, although still rigorously, and that it's unlikely that any simpler counterexample is possible. 
First, let's consider the real numbers $\mathbb{R}$ as a vector space over the field of scalars $\mathbb{Q}$. You said you're in linear algebra, so you should be able to verify that this is indeed a vector space over that field. Consider the set of real numbers $S = \{\ln(q) \mid q \in \mathbb{Q}\}$ and the subspace $W = \operatorname{Span}(S)$ of $\mathbb{R}$. Since $S$ is a spanning set for $W$ by definition, there is a subset $S' \subseteq S$ such that $S'$ is a basis for $W$. In particular, $S'$ is a linearly independent. Note that $W \neq \mathbb{R}$ since $W$ has a countable basis, and thus is a countable set, whereas $\mathbb{R}$ is uncountable. But I can then use the Axiom of Choice to extend $S'$ to a basis $B$ of $\mathbb{R}$. 
Now, by the universal properties of bases, I can choose any set function $f: B \to \mathbb{R}$, and this will have a unique extension to a linear map $\psi: \mathbb{R} \to \mathbb{R}$. First, let $b_1, b_2 \in B \setminus S', b_1 \neq b_2$, which must exist since $B \setminus S'$ is infinite (in fact, it has the same cardinality as $\mathbb{R}$). Now, I will define my set function $f$ by
$$
f(x) =
\begin{cases}
b_2,&x=b_1\\
b_1,&x=b_2\\
x,&\textrm{otherwise}
\end{cases}
$$
This then extends to a $\mathbb{Q}$-linear map $\psi$, which is a bijection of $\mathbb{R}$ since it is its own inverse function. It satisfies $\psi(x+y) = \psi(x) + \psi(y)$ for all $x,y \in \mathbb{R}$ and $\psi(b_1) = b_2, \psi(b_2) = b_1, \psi(\ln(q)) = \ln(q)$ for all $q \in \mathbb{Q}$ since the entire subspace $W$ is fixed. 
Now, we can define a map $\phi: \mathbb{R}^+ \to \mathbb{R}^+$ by $\phi(x) = e^{\psi(\ln(x))}$. $\phi$ is a bijection of $\mathbb{R}^+$ since it is a composition of bijections. It also satisfies 
$$\phi(xy) = e^{\psi(\ln(xy))} = e^{\psi(\ln(x) + \ln(y))} = e^{\psi(\ln(x)) + \psi(\ln(y))} = e^{\psi(\ln(x))} e^{\psi(\ln(y))} = \phi(x)\phi(y)$$ for all $x,y \in \mathbb{R}^+$. Note also that for any $q \in \mathbb{Q}^+$, $\phi(q) = e^{\psi(\ln(q))} = e^{\ln(q)} = q$, so $\phi$ fixes the positives rationals. BUT, $\phi(e^{b_1}) = e^{\psi(\ln(e^{b_1}))} = e^{\psi(b_1)} = e^{b_2} \neq e^{b_1}$, and so $\phi$ is not the identity function. 
The only thing we need to do now is extend $\phi$ to all of $\mathbb{R}$. So let $\varphi: \mathbb{R} \to \mathbb{R}$ be defined as follows:
$$
\varphi(x) = 
\begin{cases}
\phi(x),&x>0\\
-\phi(-x),&x<0\\
0,&x=0
\end{cases}
$$
You can check yourself that $\varphi$ is now a bijection on $\mathbb{R}$ and satisfies all the same properties that $\phi$ did, except without the restriction to positive numbers. Thus, $\varphi$ is a counterexample to your claim as stated. 
A: [Note: I mistakenly assumed that $\varphi$ conserves order, which would be implied if $\varphi(x+y)=\varphi(x)+\varphi(y)$.]
I think you almost have it.
Let $r\in\mathbb R$ and $\varepsilon>0$ be arbitrary. Choose $p,q\in\mathbb Q$ such that $p<r<q$ and $|p-r|<\varepsilon$ and $|q-r|<\varepsilon$ (which can be done since...). Now for all $x\in(p,q)$, we have 
$$
p=\varphi(p)<\varphi(x)<\varphi(q)=q
$$
This implies that $|\varphi(r)-\varphi(x)|<\varepsilon$, so $\varphi$ is continuous at $r$. 
Can you see where we use that the rationals are dense in the reals? (I.e. between any two real numbers is a rational number).
EDIT: Oh, I see now that you want to additionally prove that $\varphi(x)=x$. Well, now we know that $\varphi$ is continuous. Let $x\in\mathbb R$ be arbitrary, and choose a sequence of rational numbers $q_n\in\mathbb Q$ that converges to $x$ (which we can do because...). Now
$$
\varphi(x)
= \varphi(\lim_{n\to\infty} q_n)
= \lim \varphi(q_n)
= \lim q_n
= x
$$
